Given the function f(x) = log3(x + 4), find the value of f−1(3

First of all, find f ⁻¹ (x), or the inverse function like so:
y = log(3x + 12)
e^y = 3x + 12
3x = e^y - 12
x = 1/3 (e^y - 12)
Now, swap x and y:
f ⁻¹ (x) = y = 1/3 (e^x - 12)

Secondly and finally, substitute 3 into this eq'n:
f ⁻¹ (3) = 1/3 (e³ -12) = 2.695.....
so 2.695 (to 4 sig.figs/ 3 d.ps)
or if asked for an exact answer then it is just: 1/3 (e³ -12), which can be simplfied to 1/3e³ - 4.

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wifilethbridge wifilethbridge · Answer 2 · 2019-06-04T10:25:57+02:00

Answer:

23

Step-by-step explanation:

Given : [tex]f(x) = log_3(x + 4)[/tex]

To Find: [tex]f^{-1}(3)[/tex]

Solution:

[tex]y = log_3(x + 4)[/tex]

Replace y with x and x with y

[tex]x = log_3(y + 4)[/tex]

[tex]3^x = y + 4[/tex]

[tex]3^x-4 = y [/tex]

So, [tex]f^{-1}(x)=3^x-4[/tex]

Now substitute x= 3

[tex]f^{-1}(3)=3^(3)-4 [/tex]

[tex]f^{-1}(3)=27-4 [/tex]

[tex]f^{-1}(3)=23[/tex]

Hence the value of [tex]f^{-1}(3)[/tex] is 23